Use list-Element in XSLT

Question

So below you can see my given XML. I matched the template and I'm already in the Student-node ( xsl:template match="Class/Student" ):

<Class>
    <Heading>This is a sentence.</Heading>
    <Student>Alex</Student>
    <Student>Emilia</Student>
    <Student>John</Student>
</Class>

Now I need to get a list out of all Students and what I want to get should look like this:

<ul>
    <li>Alex</li>
    <li>Emilia</li>
    <li>John</li>
</ul>

I think I have a mistake in the way I am thinking, because my XSLT looks like this at the moment:

<xsl:template match="Class/Student">
    <ul>
        <xsl:for-each select="../Student">
            <li>
                <xsl:apply-templates/>
            </li>
        </xsl:for-each>
    </ul>
</xsl:template>

But what I actually get is:

<ul>
    <li>Alex</li>
    <li>Emilia</li>
    <li>John</li>
<ul>
<ul>
    <li>Alex</li>
    <li>Emilia</li>
    <li>John</li>
<ul>
<ul>
    <li>Alex</li>
    <li>Emilia</li>
    <li>John</li>
<ul>

I think the problem is the for-each I use but I have no idea what else I should do in this case.

Answer 1

You want one ul per Class , not per Student , so change

<xsl:template match="Class/Student">

to

<xsl:template match="Class">

Then change

    <xsl:for-each select="../Student">

to

    <xsl:for-each select="Student">

to get one li per Student child element of the Class current node.

Answer 2

As you have already made the step to use template matching with the template match="Class/Student" I would suggest to stick with that approach and simply write two templates, one for the Class elements, the other for the Student elements

  <xsl:template match="Class">
      <ul>
          <xsl:apply-templates select="Student"/>
      </ul>
  </xsl:template>

  <xsl:template match="Student">
      <li>
          <xsl:apply-templates/>
      </li>
  </xsl:template>

For more complex cases this results in cleaner and more modular code.